Question

Find the least value of ‘a’ so that $f(x) = 2x^2 - ax + 3$ is an increasing function on $[2, 4]$.

Hint:

To find the value of $a$ so that a function is increasing, check the derivative and ensure it is positive over the entire interval. For quadratic functions, focus on the lowest value in the interval.

Answer 1

We are given the function: \[ f(x) = 2x^2 - ax + 3. \] To determine the values of $a$ for which the function is increasing on $[2, 4]$, we need to find the derivative of $f(x)$: \[ f'(x) = \frac{d}{dx} \left( 2x^2 - ax + 3 \right) = 4x - a. \] For the function to be increasing on $[2, 4]$, the derivative must be positive for all $x$ in this interval, i.e., \[ f'(x)>0 \quad \text{for all } x \in [2, 4]. \] Thus, \[ 4x - a>0 \quad \text{for all } x \in [2, 4]. \] At the minimum value of $x = 2$, we have: \[ 4(2) - a>0 \quad \Rightarrow \quad 8 - a>0 \quad \Rightarrow \quad a<8. \] Therefore, the least value of $a$ such that $f(x)$ is increasing on $[2, 4]$ is \( a = 8 \).